Solving cos 3x = sin 2x using complementary angles

[misogynisticfeminist]

solve for

cos 3x=sin 2x

using complementary angles and all i got alpha=2x.

The thing is, if i got alpha 2x from sin inverse, the general formula would be

$$\theta= n \pi + (-1)^n \alpha$$

i get something like,

so,...the problem is that i will have x in the general formuala, something i don't want.

I think my way of doing it is totally wrong, can anyone help?>

btw, i know that i can use triple and double angle formula, but I've also been taught that inversing sines both sides would also work, but I've forgotten how to do it. Thanks.

 

[dextercioby]

I'd write

$$\cos 3x=\sin\left(3x+\frac{\pi}{2}\right)$$

and then

$$\sin\left(3x+\frac{\pi}{2}\right)-\sin 2x =0$$

and then i'd use a trig.identity and end up with 2 simple eqns.

Daniel.

 

[wisredz]

Well, as you sould know,

$$sin\alpha=cos(\frac{\pi}{2}-\alpha)$$

therefore, your equation would be

$$sin(2\alpha)=sin(\frac{\pi}{2}-3\alpha)$$

Try it this way...

 

[wisredz]

It seems we have posted at the same time :)

anyway I wouldn't get things into

$$\sin\left(3x+\frac{\pi}{2}\right)-\sin 2x =0$$

From what I have written there are only two possibilities. Either

$$2\alpha=\frac{\pi}{2}$$

or

$$2\alpha=\pi-(\frac{\pi}{2}-3\alpha)$$.

Maybe you meant just the same but I didn't think they could be the same...

 

[siddharth]

$$\sin 2x = \cos 3x$$

$$=> \cos (\frac{\pi}{2} - 2x) = \cos 3x$$

$$=> \frac{\pi}{2} - 2x = 2n \pi + 3x -I$$

or

$$\frac{\pi}{2} - 2x = 2n \pi - 3x - II$$

From this you can solve for the principle value of x by putting n as 0 in I.

 

[mathelord]

hey sin[x] equals cos[90-x],so in this case sin[2x] equals cos[90-2x].so equate cos[3x] to cos[90-2x].you should get 18 i guess